If $(a,c)=1$ and $b|c$, prove that $(a,b)=1$

Question

I am trying to prove this, but I don't know if I'm on the right track.

So far I've said $d=(a,c)=1$, therefore, $d|a, d|b,$ and $d|c$. I'm not sure what to do from here or if I'm even going in the right direction.

Please help!

Answer 1

If $(a,c)=1$, by Bezout identity there are $x,y\in\mathbb Z$ s.t. $$ax+cy=1.$$ Now $c=kb$ for a $k\in\mathbb Z$. Therefore $$ax+zb=1,$$ where $z=ky\in\mathbb Z$. Therefore, $(a,b)=1$ by Bezout identity.

Answer 2

Ad absurdum: assume $(a, b) = n \gneqq 1$. You know $n | a$ and $n | b$. But since $b | c$, you have $n | c$. Hence $n | (a, c)$. Conclude.

Answer 3

Since

$(a, c) = 1, \tag 1$

there exist integers $x$, $y$ such that

$ax + cy = 1; \tag 2$

now if

$(a, b) = d > 1, \tag 3$

it follows that

$d \mid a, \; d \mid b; \tag 4$

but then

$b \mid c \Longrightarrow d \mid c; \tag 5$

in the light of (1), this implies

$d \mid ax + cy \Longrightarrow d \mid 1 \Rightarrow \Leftarrow d > 1; \tag 6$

this contradiction then forces

$(a, b) = 1. \tag 7$

Answer 4

${\rm By\ Euclid}\ \ b\mid c\,\Rightarrow\, (a,b) = (a,b,c) = 1,\,$ by $\,(a,c) = 1$

${\rm Equivalently\!\!:}\ \ \underbrace{d\mid a,b\,\Rightarrow\, d\mid a,b,c}_{\large d\ \mid\ b\ \mid\ c}\,\Rightarrow\, d\mid a,c\,\Rightarrow\, d\mid 1$